A line segment from 0 0 to 0 4 forms one side of an equilateral triangle drawn in Quadrant I A Find the coordinates of the third vertex of the triangle B Find the slopes of the nonvertical line?

Answer 1

Solution (A):

Since the line segment start at origin (0, 0) and ends at (0, 4) it lies on the y-axis. So we can say that the midpoint of that line segment is (0, 2).

Since the line segment with length 4 is one of the sides of the equilateral triangle, and the lines where the two other sides of the triangle, also of length 4, lie on the first quadrant, their point of the intersection (x, y) would be the third vertex [two other vertices are (0, 0) and (0,4)].

Since in an equilateral triangle the median is also an altitude , then the line that passes through the midpoint (0, 2) and the vertex (x, y) is parallel to the x-axis. Since this horizontal line has an equation y = 2, then the y-coordinate of the third vertex would be 2; (x, 2).

The median or altitude separates the equilateral triangle into two right triangles with base length 2 and hypotenuse length 4.

From the Pythagorean Theorem, the altitude length would be 2√3 [√(42 - 22) = √12).

If from the vertex (x, y),we draw a line perpendicular to the x-axis, it intersects the x-axis at x = 2√3 (since the distance of two parallel lines intersected by two perpendicular lines, to them, is the same). Since this vertital line has an equation x = 2√3, then the x-coordinate of the third vertex would be 2√3.Thus, we can say that the third vertex would be (2√3, 2).

Or you can work using the distance formula, such as:

Since the line segment start at origin (0, 0) and ends at (0, 4) it lies on the y-axis.

Since the line segment with length 4 is one of the sides of the equilateral triangle, and the lines where the two other sides of the triangle, also of length 4, lie on the first quadrant, their point of the intersection (x, y) would be the third vertex [two other vertices are (0, 0) and (0,4)].

We know that:

the distance between the points (x, y) and (0, 0) is 4, and also

the distance between the points (x, y) and (0, 4) is 4.

By using the distance formula we have:

√[(x - 0)2 + (y - 0)2] = 4, and

√[(x - 0)2 + (y - 4)2] = 4

So that:

√[(x - 0)2 + (y - 0)2] = √[(x - 0)2 + (y - 4)2] square both sides;

x2 + y2 = x2 + y2 - 8y + 16

.

.

y = 2

Substituting 2 for y into the one of the distance formula above, we find x = 2√3

So that the third vertex would be (2√3, 2). Solution (B): The slope of the line that passes through (0, 0) and (2√3, 2) is √3. (2√3 - 0)/(2 - 0) = (2√3)/2 = √3 The slope of the line that passes through (0, 4) and (2√3, 2) is -√3. (2√3 - 0)/(2 - 4) = (2√3)/-2 = -√3